Question

The potential energy of a $1kg$ particle free to move along the x-axis is given by $U = \left( {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right)J$ . The total mechanical energy of the particle is $2J$ . Then, the maximum speed (in $m{\text{ }}{s^{ - 1}}$ ) is:
A. $\dfrac{3}{{\sqrt 2 }}$
B. $\dfrac{1}{{\sqrt 2 }}$
C. $\sqrt 2$
D. $2$

Answer 1

An object can store energy as a result of its position. This stored energy of position is referred to as potential energy. Potential energy is the stored energy of position possessed by an object.

Formula used:
$K.E. = \dfrac{1}{2}m{v^2}$
Where,
$m$ is the mass and
$v$ is the velocity.

Complete answer:
As we know, total mechanical energy is equals to the sum of kinetic energy potential energy
For maximum speed, kinetic energy should be maximum and potential energy should be minimum. That is,
$U = \left( {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right)J$ should be minimum.
For this, we have differentiate the given value,
Therefore, differentiating both the sides.

$$\dfrac{{dU(x)}}{{dx}} = \dfrac{{d\left( {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right)}}{{dx}} \\\ \Rightarrow \dfrac{{dU(x)}}{{dx}} = \dfrac{{d\left( {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right)}}{{dx}} \\\ \Rightarrow \dfrac{{dU(x)}}{{dx}} = \dfrac{{d\left( {\dfrac{{4{x^3}}}{4} - \dfrac{{2x}}{2}} \right)}}{{dx}} \\\ \Rightarrow \dfrac{{dU(x)}}{{dx}} = \left( {{x^3} - x} \right) \\\$$

For minimum potential energy, $\dfrac{{dU(x)}}{{dx}} = 0$
Therefore, $\dfrac{{dU(x)}}{{dx}} = \left( {{x^3} - x} \right) = 0$

$$\left( {{x^3} - x} \right) = 0 \\\ \Rightarrow x({x^2} - 1) = 0 \\\$$

$\Rightarrow x = 0$ or $x = \pm 1$
Now, putting the value of $x$ in $U = \left( {\dfrac{{{x^4}}}{4} - \dfrac{{{x^2}}}{2}} \right)J$ , we get,
$U = \left( { - \dfrac{1}{4}} \right)$
Now we know, total mechanical energy = Kinetic Energy + Potential Energy
Therefore,
$K.{E_{\max }} - {U_{\min }} = 2 \\\ K.{E_{\max }} = \dfrac{9}{4} \\\$
$\therefore \dfrac{1}{2}m{v^2} = \dfrac{9}{4}$
$\Rightarrow m{v^2} = \dfrac{9}{2}$ (As the given mass is $1kg$ )
$\Rightarrow {v^2} = \dfrac{9}{2} \\\ \Rightarrow v = \sqrt {\dfrac{9}{2}} \\\ \Rightarrow v = \dfrac{3}{{\sqrt 2 }}m{\text{ }}{s^{ - 1}} \\\$
So, the maximum speed of the particle is, $\dfrac{3}{{\sqrt 2 }}m{\text{ }}{s^{ - 1}}$ .
Hence, the correct option is A.

Note:
An object's kinetic energy is the energy it retains because of its motion, and potential energy is the energy an object maintains regardless of its location relative to other objects. Total mechanical energy, therefore = kinetic energy + potential energy. To answer these kinds of problems, we must note this basic principle.